If fx=∫5x8+7x6x2+2x7+12dx x≥0 and f0=0 then the value of f1=

fx=∫5x−6+7x−8x−7+x−5+22dxPut x−7+x−5+2=tDiff.5x−6+7x−8dx=−dtfx=∫1t2−dt=1t+c=x7x2+2x7+1+cGiven f0=0 ⇒ [c=0]fx=x7x2+2x7+1f1=14

If fx=∫5x8+7x6x2+2x7+12dx x≥0 and f0=0 then the value of f1=

fx=∫5x−6+7x−8x−7+x−5+22dxPut x−7+x−5+2=tDiff.5x−6+7x−8dx=−dtfx=∫1t2−dt=1t+c=x7x2+2x7+1+cGiven f0=0 ⇒ [c=0]fx=x7x2+2x7+1f1=14

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If $f (x) = \int \frac{5 x^{8} + 7 x^{6}}{{(x^{2} + 2 x^{7} + 1)}^{2}} d x$ $(x \geq 0)$ and $f (0) = 0$ then the value of $f (1) =$

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$\frac{- 1}{2}$

$\frac{- 1}{4}$

$\frac{1}{4}$

$\frac{1}{2}$

answer is C.

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Detailed Solution

$f (x) = \int \frac{5 x^{- 6} + 7 x^{- 8}}{{(x^{- 7} + x^{- 5} + 2)}^{2}} d x$
Put $x^{- 7} + x^{- 5} + 2 = t$
Diff. $(5 x^{- 6} + 7 x^{- 8}) d x = - d t$
$f (x) = \int \frac{1}{t^{2}} (- d t) = \frac{1}{t} + c = \frac{x^{7}}{x^{2} + 2 x^{7} + 1} + c$
Given $f (0) = 0 \Rightarrow [c = 0]$
$f (x) = \frac{x^{7}}{x^{2} + 2 x^{7} + 1}$
$f (1) = \frac{1}{4}$